Riemann Zeta identity

Question

[How can one prove this not using this method?1How would one go about proving $\sum _{n \ge 2} (\zeta(n) - 1)$? When you sum a Riemann $\zeta$ function you get a double sum, with the $-1$ in front of the function, the original sum for Riemann $\zeta$ starts at $n=2$ so that's where the $2$ comes from. But after that I have no idea how to prove the actual sum and the fact that its limit is $1$ and that it therefore converges to $1$. Considering partial sums is what has to be done from the looks of it but I didn't get anywhere with that.

Answer 1

Euler-Maclaurin formula gives $$\zeta(n)-1=\sum_{k=2}^\infty\frac 1{k^n}=\mathcal O\left(\frac{2^{-n}}{n-1}\right)$$ So the series converges absolutely. In fact we have $$\sum_{n=2}^\infty(\zeta(n)-1)=\sum_{n=2}^\infty\sum_{k=2}^\infty\frac 1{k^n}=\sum_{k=2}^\infty\sum_{n=2}^\infty\frac 1{k^n}=\sum_{k=2}^\infty\frac 1{k(k-1)}=1$$

Answer 2

Due to the integral representation $$ \zeta(n) = \frac{1}{(n-1)!}\int_{0}^{+\infty}\frac{x^{n-1}}{e^x-1}\,dx \tag{1}$$ we have $$ \zeta(n)-1 = \int_{0}^{+\infty}\frac{x^{n-1}}{(n-1)!}\left(\frac{1}{e^x-1}-\frac{1}{e^x}\right)\,dx \tag{2}$$ so $$\sum_{n\geq 2}\left(\zeta(n)-1\right) = \int_{0}^{+\infty}(e^x-1)\left(\frac{1}{e^x-1}-\frac{1}{e^x}\right)\,dx = \int_{0}^{+\infty}e^{-x}\,dx = \color{red}{\large 1}.\tag{3}$$ The exchange of $\sum_{n\geq 2}$ and $\int_{0}^{+\infty}$ is allowed by the dominated convergence theorem.